tion
This section is devoted to the differentiability of the minimum time function for normal linear control systems. We first give some technical results
Lemma 4.6.1. Let {rn} ⊂ R+,{ζn} ⊂ SN −1,{xn} ⊂ RN be such that
xn = M X i=1 Z rn 0 Qi(rn, t) sign gi(ζn, t)dt,
where Qi : R× R → RN, gi : RN × R → R, i = 1, · · · , M, are smooth functions. If
rn→ r, ζn→ ζ, xn → x as n → ∞ for some r ∈ R, ζ ∈ RN, x∈ RN then
x = M X i=1 Z r 0 Qi(r, t) sign gi(ζ, t)dt. Proof. Obvious.
Let S be the set of non-Lipschitz points of the minimum time function. Now we are going to define some more exceptional sets which will be useful in the sequel and study their rectifiability. All results are based on the general rectifiability statement proved in Appendix - Theorem 5.0.1.
We fix ` ∈ {1, · · · , M} and define
Σ`={ζ ∈ SN −1:∃t ∈ [0, ∞) such that hζ, eAtb`i = hζ, eAtAb`i = 0} and S` 0 = ( x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt : r ≥ 0, ζ ∈ Σ`
)
Proposition 4.6.2. S0` is countably HN −1 - rectifiable.
Proof. By Theorem 5.0.1 it is enough to show that Σ` is countablyHN −2- rectifiable. Set Σ`1 ={ζ ∈ RN : there exists t∈ [0, ∞) such that hζ, eAtb`i = hζ, eAtAb`i = 0}
and
Σ`2 ={(t, ζ) ∈ R × RN :hζ, eAtb`i = hζ, eAtAb`i = 0}.
Consider the function G : R× RN → R2 defined by
G(t, ζ) = hζ, eAtb`i, hζ, eAtAb`i , ∀(t, ζ) ∈ R × RN.
Then G is smooth and we have, for all (t, ζ), DG(t, ζ) = hζ, e AtAb `i eAtb` hζ, eAtA2b `i eAtAb` ! Since eAtb
` and eAtAb` are linearly independent for all t ∈ R, we have rankDG(t, ζ) = 2,
for all (t, ζ) ∈ R × RN. Then by Theorem 2.3.4, Σ` 2 = G
−1(0, 0) is countably
HN −1 -
rectifiable. Thus, Σ`
1 is countablyHN −1- rectifiable. It follows that Σ` is countablyHN −2
Now, let us define S` 1 = n x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt : r ≥ 0, ζ ∈ SN −1,
and hζ, eAtb`i = 0 has zeros of order ≥ 2 in [0, r]
o . Proposition 4.6.3. S`
1 is closed and countably HN −1 - rectifiable.
Proof. The rectifiability ofS1` follows the rectifiability of S0` and the fact that S1` ⊂ S0`. Let xn ∈ S1` be such that xn → x as n → ∞ for some x. Since xn ∈ S1`, there exist
rn≥ 0, tn∈ [0, rn], ζn∈ SN −1 such that xn = M X i=1 Z rn 0 eA(t−rn)b isign (hζn, eAtbii)dt and hζn, eAtnb`i = hζn, eAtnAb`i = 0.
Since xn → x, rn := T (xn) → r := T (x). By passing to subsequences, we may assume
that tn → ¯t ∈ [0, r], ζn → ζ ∈ SN −1. By Lemma 4.6.1, we have
x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt.
Moreover, we have hζ, eA¯tb`i = hζ, eA¯tAb`i = 0. This means that x ∈ S1`.
We now set S` 2 = ( x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt : r ≥ 0, ζ ∈ SN −1,hζ, b`i = 0
) .
We can see that when M = 1, the set S2` is actually the non-Lipschitz set ofT . One also can show that
Proposition 4.6.4. S`
2 is closed and countably HN −1-rectifiable.
Proof. The rectifiability of S`
2 follows Theorem 5.0.1, while the closedness can be proved
Set S` 3 = ( x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt : r ≥ 0, ζ ∈ SN −1,hζ, eArb`i = 0
) .
One can prove the following Proposition 4.6.5. S`
3 is closed and countably HN −1-rectifiable.
We now define the set
S4 = ( x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt : r ≥ 0, ζ ∈ SN −1, dim NR(r)(x)≥ 2
) .
Observe thatS4 is the set of points x∈ R where R(T (x)) is not smooth at x. We have
S4 =x ∈ R : dim NR(T (x))(x)≥ 2 .
Proposition 4.6.6. S4 is countably HN −1-rectifiable.
Proof. Set Γ ={(x, T (x)) ∈ RN +1: dim N
epi(T )(x,T (x)) ≥ 2}. Since epi(T ) has positive
reach [21], the set Γ is countablyHN −1 - rectifiable [29].
Let α : epi(T ) → RN be a mapping defined by α(x,T (x)) = x. Then α is Lipschitz. From Proposition 4.2.2, we observe thatS4 ⊂ α(Γ). It follows that S4 is countablyHN −1
- rectifiable. We set Λ =S ∪ M [ `=1 S` 1∪ M [ `=1 S` 2 ∪ M [ `=1 S` 3∪ S4.
Then, by the above results, Λ is countablyHN −1 - rectifiable. Notice that the set S 4 may
not be closed. However we will show that Λ is closed. Finally, set Ω =R \ Λ.
By the definition of Ω, we see that T is Lipschitz on Ω and that for each x ∈ Ω, the normal cone NR(T (x))(x) has only one unit vector. Therefore, by Corollary 1 in [5], T is
Observe that Ω has the following representation Ω =nx = M X i=1 Z r 0 eA(t−r)bisign hζ, eAtbii dt : r > 0, ζ ∈ SN −1,hζ, b`i 6= 0, hζ, eArb`i 6= 0 hζ, eAt
b`i has only simple zeros in [0, r], ∀` ∈ {1, · · · , M} and dim NR(r)(x) = 1
o . We have also the following characterization of Ω.
Lemma 4.6.7. A point x of the form
x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt,
for some r > 0 and ζ ∈ SN −1, belongs to Ω if and only if for each i∈ {1, · · · , M}, there
exists ki ≥ 0 such that
PM
i=1ki ≥ N − 1 and gi(ζ, t) :=hζ, e Atb
ii has ki ≥ 0 simple zeros
in (0, r), say ti j, j = 1,· · · , ki (if ki > 0), i = 1,· · · , M satisfying rank n eAtijb i : j = 1,· · · , ki, i = 1,· · · , M o = N− 1. (4.6.1) and gi(ζ, t) has no more zeros in [0, r].
Proof. Assume that
x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt,
for some r > 0 and ζ ∈ SN −1 and that x ∈ Ω. Assume also that gi(ζ, ·) =hζ, eA·bii has
ki ≥ 0 zeros in [0, r]. Since x ∈ Ω, there exists ` ∈ {1, · · · , M} such that k` > 0 and
gi(ζ, ·) has only ki simple zeros in (0, r) and has no more zeros in [0, r].
Set I = {i ∈ {1, · · · , M} : ki > 0}. For each i ∈ I, let 0 < ti1 <· · · < tiki < r be zeros
of gi(ζ, ·) in [0, r]. Since tij is simple zero of gi(ζ, ·), we havehζ, eAt
i jAb
ii 6= 0 for all i ∈ I
and j = 1,· · · , ki. Then for each i ∈ I, there exist open neighborhoods V in SN −1 of ζ
and Iji in [0, r] of tij such that Iji∩ Imi =∅ if j 6= m and hη, eAtAbii 6= 0 for all η ∈ V and
t∈ Iji, j = 1,· · · , ki.
Set Ii = [0, r]\ ∪ki
j=1Iji for each i ∈ I. Then Ii is closed. Set σi = mint∈Ii|hζ, eAtbii|
and σ = min{σi : i∈ I}. Observe that σ > 0. By the continuity, we can choose V and Iji
such that |hη, eAt bii| ≥ σ 2 > 0, ∀η ∈ V, ∀t ∈ I i .
Suppose that (4.6.1) fails i.e., rank{eAtijb
i : 1 ≤ j ≤ ki, i ∈ I} ≤ N − 2. Then there
exists ¯ζ ∈ SN −1such that ¯ζ, ζ are linearly independent andh¯ζ, eAtijb
ii = 0 for all 1 ≤ j ≤ ki
and for all i∈ I. Choose λ > 0 sufficiently small such that ζ1 := ζ + λ¯ζ ∈ V . Then ζ and
ζ1 are linearly independent andhζ1, eAt
i jb
ii = 0 for all j ∈ {1, · · · , ki} and |hζ1, eAtbii| > 0,
for all t ∈ Ii, i ∈ I. Since hζ1, eAtAbii 6= 0 for all t ∈ Iji, we observe that tij are all zeros
of hζ1, eA·bii in [0, r] and they are simple zeros. Therefore
x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt = M
X
i=1
Z r
0
eA(t−r)bisign (hζ1, eAtbii)dt.
Since ζ, ζ1 are linearly independent, dim NR(r)(x)≥ 2. This leads to a contradiction with
x∈ Ω. The other implication is obvious. The proof is complete. Proposition 4.6.8. Ω is open.
Proof. Assume to the contrary that Ω is not open. Then there exist a point x∈ Ω and a sequence {xk} ⊂ Λ such that
xk → x as k → ∞.
For eack k, there exists ζk∈ SN −1 such that
xk = M X i=1 Z rk 0 eA(t−rk)b isign (hζk, eAtbii)dt, where rk = T (xk).
Since ζk ∈ SN −1, we may assume that ζk → ζ ∈ SN −1 as k→ ∞. On the other hand,
since xk→ x, rk → r := T (x). Therefore x = M X i=1 Z r 0
eA(t−r)bisign (hζ, eAtbii)dt.
By Lemma 4.6.7, hζ, eA·b
ii has only ki ≥ 0 simple zeros in (0, r) and has no other zeros
in [0, r]. We denote by A the set of i ∈ {1, · · · , M} such that ki ≥ 1. For i ∈ A, let
0 < ti
1 <· · · , tiki < r be all zeros of hζ, e
A·b
ii in [0, r], then one has
X i∈A ki ≥ N − 1 and rank n eAtijb i : 1≤ j ≤ ki, i∈ A o = N − 1.
Since ti1,· · · , tiki are simple zeros of hζ, eA·bii in [0, r], we can find open neighborhoods V
in SN −1 of ζ and Iji in [0, r] of tij, 1 ≤ j ≤ ki and I0 of r such that Iji∩ I`i = ∅ for j 6= `
and Iji ∩ I0 = ∅ and for all η ∈ V the equation hη, eAtbii = 0 has only one simple zero
in Iji and has no more zeros in [0, ¯r], for all ¯r ∈ I0 with i∈ A and further the equation hη, eAtb
mi = 0 has no zero in [0, ¯r] for all ¯r ∈ I0 with m∈ {1, · · · , M} \ A.
For any η ∈ V , let sij ∈ Iji be the zeros of hη, eAtbii, 1 ≤ j ≤ ki, i ∈ A, then by
choosing V, Iji, I0 small enough, we obtain that rankneAsijb
i : 1≤ j ≤ ki, i∈ A
o
= N − 1. It follows from Lemma 4.6.7 that
¯ x := M X i=1 Z r¯ 0
eA(t−¯r)bisign (hη, eAtbii)dt ∈ Ω,
for all η ∈ V and ¯r ∈ I0. Therefore xk ∈ Ω for k sufficiently large. This contradiction
concludes that Ω is open.
From the above results, we observe that
Theorem 4.6.9. The minimum time function is of class C1 in an open set Ω whose
complement in the reachable set is countably HN −1 rectifiable.
We end this section with a proposition which can be seen as a propagation property of the differentiability of the minimum time function along optimal trajectories
Proposition 4.6.10. Let x6= 0 and y(·) be the optimal trajectory for x. Let r ∈ (0, T (x)) be such that T is differentiable at y(r). Then T is differentiable at y(s) for all s ∈ [0, r]. Proof. Since T is differentiable at y(r), the normal cone NR(T (y(r)))(y(r)) has only one
unit vector. Hence NR(T (y(s)))(y(s)) also has only one unit vector (see, e.g., exersice 15.1
[34]). Observe thatT is Lipschitz at y(s) for all s ∈ [0, r]. Indeed, if T is not Lipschitz at y(s) for some s∈ [0, r]. Then T is not Lipschitz at y(t) for all t ∈ [s, T (x)]. Hence T is not Lipschitz at y(r) which is a contradiction. Thus, by Corollary 1 [5],T is differentiable at y(s) for all s∈ [0, r].
Remark 4.6.11. Notice that in Proposition 4.6.10, T may not be differentiable at y(s) for some s∈ (r, T (x)].